T2 = I (identity permutation, e.g the permutation such that I(i) = i for all i = 1,...n)

A permutation p of the integers {1, . . . n} is denoted by

[1 .. . .. n]

[p(1).. p(n)]

So that, for example:

[1 2 3]

[2 1 3]

denotes the permutation p such that p(1) = 2, p(2) = 1, and p(3) = 3.

Now, let's look at EVEN and ODD permutations:

Let P_n denote the polynomial of n variables x1, x2……xn which is the product of all the factors x_i … xj with i < j. That is:

P_n(x1, x2... . xn) = P(x_i .. x_j)

The symmetric group S(n) acts on the polynomial P_n by permuting the variables. For p C S(n) we have:

P_n( x_p(1), x_p(2). . .x_p(n)) = (sgn p) P_n (x_1, x_2…..x_n)

where sgn p =

**. If the sign is positive then p is called an even permutation, if the sign is negative then p is called an odd permutation. Thus: the product of two even or two odd permutations is even. The product of an even and an odd permutation is odd.**

__+__1Back to transpositions! We just saw:

[1 2 3]

[2 1 3]

The above permutation is actually a transposition 2 <-> 1 (leaving 3 fixed). Now, let p' be the permutation:

[1 2 3]

[3 1 2]

Then pp' is the permutation such that:

pp'(1) = p(p'(1)) = p(3) = 3

pp'(2) = p(p'(2)) = p(1) = 2

pp'(3) = p(p'(3)) = p(2) = 1

It isn’t difficult to ascertain that: sgn (ps) = (sgn p) (sgn s) so that we may write:

pp' =

[1 2 3]

[3 2 1]

Now, find the inverse p^-1 of the above. (Note: the inverse permutation, denoted by p^-1 is defined as the map: p^-1 : Zn -> Zn), Since p'(1) = 3, then p ^-1(3) = 1

Since p'(2) = 1 then p^ -1(1) = 2

Since p'(3) = 2 then p^ -1(2) = 3

Therefore: p^-1 =

[1 2 3]

[ 2 3 1]

Problem: Express p =

[1 2 3 4]

[2 3 1 4]

as the product of transpositions, and determine the sign (+1 or -1) of the resulting end permutation.

Let T1 be the transposition 2 <-> 1 leaving 3, 4 fixed, so:

T1 p =

[1 2 3 4]

[1 3 2 4]

Let T2 be the transposition 2 <-> 3 leaving 1, 4 fixed, so:

T2 T1 p =

[1 2 3 4]

[1 2 3 4]

Then:

T2 T1 p = I (identity)

TWO transpositions (T1, T2) operated on p, so that the sign of the resulting permutation (to reach identity) is +1. The permutation is therefore even.

**:**

*A Note on Disjoint Permutations*Expressing a permutation as a product of disjoint cyclic permutations is not hard at all, once one gets the hang of it. The key is to “cycle through” the mapping in the original to yield the different disjoint cycles, taking care to stop when the end element leads to a number (on the top of the original) that repeats. For example:

Express as disjoint permutations:

[1 2 3 4 5 6 7]

[4 5 6 7 3 1 2]

Solution: 1 goes into 4, 4 goes into 7, 7 goes into 2 – STOP! (Since 2 commences new cycle in next top position). So first disjoint cycle is: (1, 4, 7, 2).

Now, 2 goes into 5, 5 goes into 3, STOP! (3 repeats) So cycle is: (2, 5, 3). Then 3 goes into 6, and 6 into 1. Stop.

Answer: (1, 4, 7, 2)(2, 5, 3)(6).

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